# Bad proof that if $a + b + ab = 2020$ then $a+b=88$

Can you prove this:

Let $a,b \in \mathbb{N}$. If $a + b + ab = 2020$ then $a+b=88$.

This is the attempt given:

$\frac{2020-88}{a b}=1$

$a+b=88$

Substituting for $b$ using the $2$nd equation.

$2020-88 = a (88-a)$

That is a quadratic that is easily solved and gets $a = 42$ and $a = 46$.

So we have $a = 42$ and $b = 46$ or $a = 46$ and $b = 42$.

Was the question answered? A valid proof?

• I do not see in your proof how you get to $\frac{2020 - 88}{ab} = 1$. Can you explain this step?? Haven't you just assumed $a + b = 88$ to get to this step?
– 6005
May 10, 2014 at 17:53
• Yes, I have, that is the problem. May 10, 2014 at 17:55
• Are you trying to prove $a + b = 88$, or is that a "given" from which you are trying to derive the values $a, b$. I ask because you state the proposition as an "if, then" statement. To prove it, you can only assume the "if" condition, in order to prove the then conclusion. If that's what you're trying to do, you cannot assume (or use) what it is you are to prove. May 10, 2014 at 17:56
• I do not know, the question is exactly as asked, what do you think? May 10, 2014 at 17:57
• What you have proved is the other way around: if $a+b+ab=2020$ and $a+b=88$, then $a,b\in\mathbb N$. May 11, 2014 at 16:17

$a+b+ab=2020 \iff (a+1)(b+1)=2021=47 \cdot 43$

• That was one other proof provided. May 10, 2014 at 18:00
• @bobbym See my answer for how to view this as a generalization of completing squares to completing products, a general method which works for any Diophantine equation of this form. May 10, 2014 at 20:56
• @Bill Dubuque Looking at it now. May 11, 2014 at 2:38

Your proof is not valid, assuming I understood you correctly.

This is the attempt given:

$$\frac{2020-88}{a b}=1$$

$$a+b=88$$

The problem is you started out by assuming $$a + b = 88$$. You just assumed what you wanted to prove!

Here is the question again:

Can you prove this: if $$a,b$$ are positive integers, and if $$a + b + ab = 2020$$, then $$a+b=88$$?

Notice that the fact that $$a, b$$ are positive integers is very necessary here. Otherwise, you could pick any rational $$a,b$$ with $$(a + 1)(b + 1) = 2021$$ (in this case $$a$$ can be any rational number) and you would have a solution. In general $$a + b \ne 88$$ if $$a,b$$ are rational.

Thus your proof should be suspicious: you haven't used any properties about integers, as far as I can see. Your proof would also conclude that $$a + b = 88$$ if $$a,b$$ are rational, and this is not a true result! Therefore, your proof cannot be valid.

The correct proof is as leticia gives: write $$(a + 1)(b+1) = 2021 = 47 \cdot 43$$, and use prime factorization -- a property of positive integers -- to derive your result.

• Very nice answer. +1 May 10, 2014 at 20:50

Key Idea $\$ Completing a square generalizes to two-variables $\,x,y\,$ to completing a product

$$\begin{eqnarray} && xy + bx + cy\\ &=\,& x(y\!+\!b) + cy \\ &=\,& x(y\!+\!b) + c(y\!+\!b)-cb \\ &=\,& \qquad\, (x\!+\!c)(y\!+\!b) - cb \end{eqnarray}\qquad\qquad\quad$$

${\rm So}\ \ b=1=c\ \ {\rm yields}\ \ n = xy\!+\!x\!+\!y = (x\!+\!1)(y\!+\!1)-1\!\iff\! n\!+\!1 = (x\!+\!1)(y\!+\!1)\$ which has solutions $\,x,y = j\!-\!1,k\!-\!1\,$ for each factorization $\,n\!+\!1 = jk.\,$

Remark $\$ In the same way we can show that if $\,a\ne 0\,$ then $$\ axy\! +\!bx\!+\!cy = n\!\! \overset{\ \times\,a}\iff\! (ax\!+\!c)(ay\!+\!b) = an\!+\!cb$$

Alternatively one can mechanically derive the above from the first special case $(a = 1)$ using the $\rm AC$-method as described here.

• Nice explanation.
– rugi
Feb 29, 2016 at 5:48

You demonstrated the that both can co-exist and have a solution:

$\exists! (a, b) \in ℕ$ such as $a+b=88$ and $a+b+ab=2020$

Now if you can prove the uniqueness of the second part

$\exists! (a, b) \in ℕ$ such as $a+b+ab=2020$

That would be sufficient. (however, proving the uniqueness of the solution is quite likely to result in actually finding that solution, rendering the first part of the reasoning useless)